Suppose that $f(x)$ is a function such that
\[f(xy) + x = xf(y) + f(x)\]for all real numbers $x$ and $y.$  If $f(-1) = 5$ then compute $f(-1001).$
Solution: Setting $y = 0$ in the given functional equation, we get
\[f(0) + x = xf(0) + f(x),\]so $f(x) = (1 - f(0))x + f(0).$  This tells us that $f(x)$ is a linear function of the form $f(x) = mx + b.$  Since $f(-1) = 5,$ $5 = -m + b,$ so $b = m + 5,$ and
\[f(x) = mx + m + 5.\]Substituting this into the given functional equation, we get
\[mxy + m + 5 + x = x(my + m + 5) + mx + m + 5.\]This simplifies to $2mx = -4x.$  For this to hold for all $x,$ we must have $m = -2.$

Then $f(x) = -2x + 3.$  In particular, $f(-1001) = \boxed{2005}.$